Find $\det(\mathrm{I} - 2\mathrm{P})$ where $\mathrm{P}\mathrm{P} = \mathrm{P}$ is a projection matrix

Question

Let $\mathrm{P}$ be a well-defined $n\times n$ projection matrix and $\mathrm{I}$ be the identity matrix. What is the determinant of the following reflection matrix? $$ \mathrm{R} = \mathrm{I} - 2\mathrm{P} $$

Important: The projection matrix in my case is always defined as $\mathrm{P} = \mathrm{A}^\top(\mathrm{A}\mathrm{A}^\top)^{-1}\mathrm{A}$ where $\mathrm{A}$ is a full row-rank $m\times n$ matrix with $n > m$.

Attempt following comments

The determinant of $\mathrm{R}$ is the product of its eigenvalues $$ \det(\mathrm{R}) = \prod_{i=1}^n \lambda_i. $$ Any non-zero vector $v\in\mathbb{R}^n$ such that $$ \mathrm{R}v = \lambda v $$ is an eigenvector with eigenvalue $\lambda$. This can be written $$ (\mathrm{R} - \lambda \mathrm{I}) v = 0 $$ which has a non-zero solution only if $\det((1-\lambda)\mathrm{I} - 2\mathrm{P}) = 0$. However I am not sure how to show this.

Answer 1

The eigenvalues of $P$ are $0$ and $1$, with the multiplicity of $1$ equal to the rank of $P$. It follows that the eigenvalues of $I - 2 P$ are $1 - 2(0) = 1$ and $1 - 1(1) = -1$, with the multiplicity of $-1$ equal to the rank of $P$.

Because the determinant of a matrix is the product of its eigenvalues, we have $$ \det(I - 2P) = 1^{n - \operatorname{rank}(P)} (-1)^{\operatorname{rank}(P)} = (-1)^{\operatorname{rank}(P)}. $$

Answer 2

let $R=I-2P, \;P^2=P$

$det(PR)=det(P(I-2P))\\=det(P-2P^2)=det(P-2P)\\=det(-P)=det(P)det(R)$

now $det(-P)=det(-IP)\\=det(-I)det(P)=-det(P)$

thus:

$det(R)=(-1)^n$.